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Polar Form and DMT.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \begin{document} {\large Complex Number \--- Polar Form and De Moivre's Theorem} \begin{align*} \text{Modulus:}\quad&|\bar{z}|=|-z|=|z|\\ &|kz|=k|z|\:,\quad\text{where }k\in\mathbb{R+}\:.\\ &z\bar{z}=|z|^2\\ &\frac{\:z\:}{\overline{z}}=\cos 2\theta+i\sin 2\theta\\ &|z_1\cdot z_2|=|z_1|\cdot|z_2|\\ &\left|\frac{z_1}{z_2}\right|=\frac{\:|z_1|\:}{|z_2|}\\ &|z_1+z_2|^2+|z_1-z_2|^2=2\left[|z_1|^2+|z_2|^2\right]\qquad\big(\text{\small$(z_1+z_2)(\overline{z_1}+\overline{z_2})+(z_1-z_2)(\overline{z_1}-\overline{z_2})=2\:z_1\:\overline{z_1}+2\:z_2\:\overline{z_2}$}\big)\\ &|z_1+z_2|\leq|z_1|+|z_2|\\ &\big||z_1|-|z_2|\big|\leq|z_1+z_2|\leq|z_1|+|z_2|\qquad\big(\text{\small$|z_1|=|(z_1+z_2)+(-z_2)|\leq|z_1+z_2|+|-z_2|,\:|z_1|-|z_2|\leq|z_1+z_2|$}\big)\\ &\text{Given }z_1\neq 0\text{ and }z_2\neq 0\:,\;|z_1+z_2|=|z_1|+|z_2|\;\Leftrightarrow\;\frac{z_1}{z_2}\in\mathbb{R}\;\Leftrightarrow\;\text{$O$, $P_1$ and $P_2$ are collinear.}\\ \\ \text{Arguments:}\quad&\text{Principal argument value is in either }(-\pi,\pi]\text{ (to match the range of $\tan^{-1}$) or }[0,2\pi)\text{ (to be positive).}\\ &\text{Here we denote the principal argument value of complex number $z$ with $arg(z)\in(-\pi,\pi]$ .}\\ &\text{If a formula may result in an argument value out of the $(-\pi,\pi]$ domain, $\text{mod}\:2\pi$ is required.}\\ &\arg(z)=0\:,\quad\text{If $z$ is purely real and $z\geq 0$ .}\\ &\arg(z)=\pi\:,\quad\text{If $z$ is purely real and $z<0$ .}\\ &\arg(z)=\tfrac{\:\pi\:}{2}\:,\quad\text{If $z$ is purely imaginary and $z=ki$ where $k\in\mathbb{R+}$ .}\\ &\arg(z)=-\tfrac{\:\pi\:}{2}\:,\quad\text{If $z$ is purely imaginary and $z=ki$ where $k\in\mathbb{R-}$ .}\\ &\arg(-z)=\arg(z)-\pi\:,\quad\text{where }\arg(z)\neq 0\\ &\arg(\bar{z})=-\arg(z)\:,\quad\text{where }\arg(z)\neq\pi\\ &\arg(z_1\cdot z_2)\equiv arg(z_1)+arg(z_2)\quad(\text{mod}\:2\pi)\qquad\Big(\text{e.g. }\arg(\text{cis}(\pi)\cdot\text{cis}(\tfrac{\:\pi\:}{2}))\equiv\tfrac{3\pi}{2}\equiv-\tfrac{\:1\:}{\pi}\quad(\text{mod}\:2\pi)\Big)\\ &\arg(z^n)\equiv n\:\arg(z)\quad(\text{mod}\:2\pi)\\ &\arg\left(\frac{\:1\:}{z}\right)=-\arg(z)\:,\quad\text{where }\arg(z)\neq\pi\\ &\arg\left(\frac{\:z_1\:}{z_2}\right)\equiv\arg(z_1)-\arg(z_2)\quad(\text{mod}\:2\pi)\\ \\ \text{De Moivre's Theorem:}\quad&[r_1(\cos\theta_1+i\sin\theta_1)]\cdot[r_2(\cos\theta_2+i\sin\theta_2)]=(r_1\cdot r_2)[\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)]\\ &\text{Proof:}\quad LHS=(r_1\cdot r_2)[(\cos\theta_1\:\cos\theta_2-\sin\theta_1\:\sin\theta_2)+i(\sin\theta_1\:\cos\theta_2+\cos\theta_1\:\sin\theta_2)]\\ &=(r_1\cdot r_2)[\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)]=RHS\\ \\ &\overline{(\cos\theta+i\sin\theta)}=\cos(-\theta)+i\sin(-\theta)\\ &\tfrac{1}{\cos\theta+i\sin\theta}=\cos(-\theta)+i\sin(-\theta)\\ \\ &(\cos\theta+i\sin\theta)^0=\cos 0+i\sin 0=1\\ \\ &\text{If $n$ and $m$ are positive integers,}\\ &(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta\:,\qquad (\cos\theta+i\sin\theta)^{-n}=\cos(-n\theta)+i\sin(-n\theta)\\ &(\cos\theta+i\sin\theta)^{\frac{n}{m}}=\cos\tfrac{n}{m}\theta+i\sin\tfrac{n}{m}\theta\:,\qquad (\cos\theta+i\sin\theta)^{-\frac{n}{m}}=\cos(-\tfrac{n}{m}\theta)+i\sin(-\tfrac{n}{m}\theta)\\ \end{align*} \end{document}